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Zorluk: OrtaPhotoelectric Effect and Work Function

A photosensitive metal plate with a work function of 2.3 eV2.3\text{ eV} is illuminated by light of frequency 8.0×1014 Hz8.0 \times 10^{14}\text{ Hz}. If the intensity of the light source is doubled while maintaining the same frequency, what will be the new maximum kinetic energy of the emitted photoelectrons? (h=6.6×10−34 J sh = 6.6 \times 10^{-34}\text{ J s}, 1 eV=1.6×10−19 J1\text{ eV} = 1.6 \times 10^{-19}\text{ J})

  1. remain unchanged at 1.0 eV1.0\text{ eV}Cevap
  2. B
    double to 2.0 eV2.0\text{ eV}
  3. C
    quadruple to 4.0 eV4.0\text{ eV}
  4. D
    increase to 3.3 eV3.3\text{ eV}

Cevap

The maximum kinetic energy will remain unchanged at 1.0 eV1.0\text{ eV}.
According to Einstein's photoelectric equation, Kmax=hf−W0K_{\text{max}} = hf - W_0. The maximum kinetic energy of an emitted photoelectron depends exclusively on the frequency of the incident photons and the work function of the target metal. Increasing the light intensity increases the rate of photon arrival and thus the rate of photoelectron emission, but it does not change the energy of individual photons. Therefore, the maximum kinetic energy remains constant at 1.0 eV1.0\text{ eV}.

Adım Adım Çözüm

1
Calculate the energy of the incident photons in Joules and convert to electron-volts (eV)
E=hf=6.6×10−34 J s×8.0×1014 Hz=5.28×10−19 JE = hf = 6.6 \times 10^{-34}\text{ J s} \times 8.0 \times 10^{14}\text{ Hz} = 5.28 \times 10^{-19}\text{ J}. In eV: E=5.28×10−191.6×10−19=3.3 eVE = \frac{5.28 \times 10^{-19}}{1.6 \times 10^{-19}} = 3.3\text{ eV}.
Einstein's photoelectric equation requires comparing photon energy with the work function of the metal.
2
Calculate the maximum kinetic energy of the photoelectrons using Einstein's photoelectric equation
Kmax=E−W0=3.3 eV−2.3 eV=1.0 eVK_{\text{max}} = E - W_0 = 3.3\text{ eV} - 2.3\text{ eV} = 1.0\text{ eV}.
The maximum kinetic energy is the surplus energy after overcoming the metal's work function.
3
Analyze the effect of doubling light intensity at constant frequency
Doubling intensity increases the photon flux (number of photons per second), thereby increasing emission current, but leaves individual photon energy and KmaxK_{\text{max}} completely unchanged at 1.0 eV1.0\text{ eV}.
Kinetic energy of individual photoelectrons depends strictly on photon frequency, not beam intensity.

Anahtar Kavram

Independence of photoelectron kinetic energy from light intensity
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Konular

Photoelectric Effect and Work Function

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